integral from-1 to pi-1 of sin(1+x)

Lösung

\int_{- 1}^{π - 1} sin (1 + x) d x

2

Schritte zur Lösung

\int_{- 1}^{π - 1} sin (1 + x) d x

Wende U-Substitution an

= \int_{0}^{π} sin (u) d u

Nutze das gemeinsame Integral :

\int sin (u) d u = - cos (u)

= [- cos (u)]_{0}^{π}

Berechne die Grenzen:

2

= 2

\int_{0}^{2} x^{2} + 2 d x

\int_{0}^{z^{2}} z d z

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\int_{5}^{11} \frac{y}{y ^{2} - y - 2} d y